Question: The region shown is bounded by the arcs of circles having radius 4 units, having a central angle measure of 60 degrees and intersecting at points of tangency. The area of the region can be expressed in the form $a\sqrt{b}+c\pi$ square units, where $\sqrt{b}$ is a radical in simplest form. What is the value of $a + b + c$?
[asy]
size(150);
draw(arc((-2,0),2,0,60));
draw(arc((0,3.464),2,-60,-120));
draw(arc((2,0),2,120,180));
[/asy]
Answer: Consider point $A$ at the center of the diagram. Drawing in lines as shown below divides the region into 3 parts with equal areas. Because the full circle around point $A$ is divided into 3 angles of equal measure, each of these angles is 120 degrees in measure.
[asy]
size(150);
pair A, B, C, D;
A=(0,1.155);
B=(0,0);
C=(-1,1.732);
D=(1,1.732);
draw(arc((-2,0),2,0,60));
draw(arc((0,3.464),2,-60,-120));
draw(arc((2,0),2,120,180));
dot(A);
label("A", A, N);
draw(A--B);
draw(A--C);
draw(A--D);
[/asy] Now consider a circle of radius 4 inscribed inside a regular hexagon:
[asy]
size(150);
pair O, A, B, C, D, E, F, M;
O=(0,0);
A=(-4.619,0);
B=(-2.309,4);
C=(2.309,4);
D=(4.619,0);
E=(2.309,-4);
F=(-2.309,-4);
M=(A+B)/2;
draw(circle(O,4));
draw(A--B--C--D--E--F--A);
label("A", A, W);
label("B", B, NW);
label("O", O, SE);
label("C", C, NE);
label("D", D, E);
label("E", E, SE);
label("F", F, SW);
label("M", M, NW);
draw(A--O);
draw(B--O);
draw(M--O);
label("$4$", 3M/4, NE);
[/asy] Now, the pieces of area inside the hexagon but outside the circle are identical to the pieces of area the original region was divided into. There were 3 pieces in the original diagram, but there are 6 in the hexagon picture. Thus, the area of the original region is the half the area inside the hexagon but outside the circle.

Because $ABO$ is equilateral, $BMO$ is a 30-60-90 right triangle, so $BM=\frac{4}{\sqrt{3}}$. Thus, the side length of the equilateral triangle is $AB=2BM=\frac{8}{\sqrt{3}}$. Now we know the base $AB$ and the height $MO$ so we can find the area of triangle $ABO$ to be $\frac{1}{2} \cdot \frac{8}{\sqrt{3}} \cdot 4=\frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$. The entirety of hexagon $ABCDEF$ can be divided into 6 such triangles, so the area of $ABCDEF$ is $\frac{16\sqrt{3}}{3} \cdot 6 = 32\sqrt{3}$. The area of the circle is $\pi 4^2=16\pi$. Thus, the area inside the heagon but outside the circle is $32\sqrt{3}-16\pi$. Thus, the area of the original region is $\frac{32\sqrt{3}-16\pi}{2}=16\sqrt{3}-8\pi$.

Now we have $a=16$, $b=3$ and $c=-8$. Adding, we get $16+3+(-8)=\boxed{11}$.